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Calculating Empirical and Molecular Formulas

Calculating Empirical and Molecular Formulas. What is an Empirical Formula? The formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula. What is a Molecular Formula?

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Calculating Empirical and Molecular Formulas

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  1. Calculating Empirical and Molecular Formulas What is an Empirical Formula? The formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula. What is a Molecular Formula? It is the formula of a compound in which the subscripts give the actual number of each element in the formula.

  2. Calculating Empirical Formula • What if a percent composition is given. How do we find the empirical formula? • Example: Chemical analysis of a liquid shows that it is 60.00% C, 13.40% H, and 26.60% O by mass. Calculate the empirical formula of this substance. • Step 1: Convert to grams • Assume you have a 100.00 g sample, and convert percentages to grams. • 60.00% Carbon is the same as 60.00 g of C • 13.40% of Hydrogen is the same as 13.40 g of H • 26.60% of Oxygen is the same as 26.60 g of O

  3. Step 2: Convert grams to molesStep 3: Divide each by the smallest decimal (in moles) to get whole numbers.* These numbers will be the subscripts. 1 mole C 60.00 g C = 5.00 mole C = 3.01 mol C ÷ 1.66 12.01 g C Round those answers to the nearest whole number and use them as the Subscripts on the final Empirical Formula. 13.40 g H 1 mole H = 13.27mole H = 7.99 mol H ÷ 1.66 1.01 g H 26.60 g O 1 mole O = 1.66 mole O ÷ 1.66 = 1.00 mol C 16.00 g O THE EMPIRICAL FORMULA WOULD BE: C3H8O1 C3H8O Of the three calculated numbers, this is the smallest decimal in moles, so you divide all the other decimals and itself by that number.

  4. Example 2 for Empirical Formulas • A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? • Step 1: Convert to grams : 72.2 % Mg = 72.2 g Mg 27.8 % N = 27.8 g N • Step 2: Convert grams to moles 1 mole Mg 72.2 g Mg = 1.49 mole Mg = 2.97 mole Mg 24.31 g Mg 27.9 g N 1 mole N ÷ 1.99 = 1.99 mole N 14.01 g N Step 3: Divide each by the smallest decimal (in moles) to get whole numbers. • Still not whole numbers. We have to get rid of the 1.49. If we multiply by 1 we get the same number of course. So, lets multiply by 2 and see what happens. Remember whatever we do to one side we do it to the other. 1.00 mole N x 2 = 2 mole N THE EMPIRICAL FORMULA WOULD BE: Mg3N2

  5. How do we determine the molecular formula from the empirical formula? • Example: The empirical formula for a compound isP2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound. • Step 1: Find the molar mass of the empirical formula. • 2 mol P x 30.97 g = 61.94 g P • 5 mol O x 16.00g = + 80.00 g O 141.94 g/mol

  6. Step 2: Molar mass of a compound divided by the molar mass of empirical formula. Experimental molar mass of compound Molar mass of empirical formula 284 g/mol 141.94 g/ mol = 2.00 YOUR MOLECULAR FORMULA IS: P4O10

  7. Assignment Determine the Empirical Formula for each of the following. Use the calculated empirical formula to calculate the molecular formula. • A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molecular formulas. • A compound has a molar mass of 456.18 g/mol and consists of 3.0 g of Fe and 4.81g of S. Calculate both empirical and molecular formulas • A compound consists of 36.48% Na, 25.41% S, and 38.11% O. It has a molar mass of 252.10 g/mol. Calculate both empirical and molecular formulas.

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